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For which values of a,b,c is this true?

  1. Nov 9, 2012 #1
    1. The problem statement, all variables and given/known data

    For which values of [itex]a,b,c\in\mathbb{C}[/itex] is the following equation true? [tex]a(x+t)^2 + b(x+t) + c = \alpha(ax^2 + bx + c)[/tex] where [itex]\alpha[/itex] is some scalar.

    3. The attempt at a solution

    How do I go about this?
     
  2. jcsd
  3. Nov 9, 2012 #2

    haruspex

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    I don't understand the question. What roles do x and t play here? Is the equation to be true for all x and t? Some x and t? ....
     
  4. Nov 9, 2012 #3
    [itex]x[/itex] is a variable and [itex]t\in\mathbb{R}[/itex] and [itex]\alpha[/itex] is a fixed constant. We want the equation to be true for all [itex]t[/itex].

    If you equate coefficients you get:

    [itex]a=\alpha a[/itex]
    [itex]2ta+b = \alpha b[/itex]
    [itex]at^2 + bt + c = \alpha c[/itex]

    For what values of a, b and c are these true?
     
  5. Nov 9, 2012 #4

    haruspex

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    You mean for all x, I assume. Is alpha real?
    Try considering α=1, α≠1 separately. (That's alpha, not a.)
     
  6. Nov 9, 2012 #5
    If [itex]\alpha =1[/itex] then [itex]a=0[/itex] and [itex]b=0[/itex]

    If [itex]\alpha \neq 1[/itex] then the first equation implies [itex]a(1-\alpha) = 0[/itex] so [itex]a=0[/itex] since [itex]\alpha \neq 1[/itex].

    Subbing [itex]a=0[/itex] into the second equation gives [itex]b=\alpha b[/itex] so [itex]b(1-\alpha)=0[/itex] so [itex]b=0[/itex] since [itex]\alpha \neq 1[/itex].

    Subbing [itex]a=0, b=0[/itex] in the third equation gives [itex]c=0[/itex]
     
  7. Nov 9, 2012 #6

    haruspex

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    Unless t = 0.
     
  8. Nov 10, 2012 #7
    So, assuming [itex]t\neq 0[/itex] (for if [itex]t=0[/itex] the equation is trivially true), I can conclude that the equation will be true for [itex]\alpha =1[/itex] for all [itex]c\in\mathbb{C}, a=0, b=0[/itex] and when [itex]\alpha \neq 1[/itex] it will only be true for [itex]a,b,c=0[/itex]?

    In other words, whatever the value of [itex]\alpha[/itex], the equation will be true for all [itex]c\in\mathbb{C}[/itex] with [itex]a,b=0[/itex] so the polynomial involved [itex]p(x) = ax^2 + bx + c[/itex] must be constant; i.e. [itex]p(x) = c[/itex].
     
    Last edited: Nov 10, 2012
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